Part Number: TXB0104
For TXB0104QRU, it’s a 4-bit level shifter. Our application VCCA=VCCB=1.8V. What’s the recommendation for OE pin pull up resistor value ? 100K or 10K ?
If I calculate as 1uA, 1.8/1uA=1.8M ohm. Could you please comment on this？ Thanks a lot.
A 1.8 MΩ resistor with a current of 1 µA would have a voltage drop of 1.8 V volt, and force the pin low. 10 kΩ or 100 kΩ would be fine.
If you never want OE to be low, you can connect it directly to VCC.
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In reply to Clemens Ladisch:
Thanks for your reply. I don't fully understand the working principal of this small circuit. Does the OE pin internal circuit is same as figure shown below? If it is same as this circuit. When Vin pull high which will make low side MOS close and the output logic is low? Could you please help me to understand this working principal? Thanks a lot!
In reply to Zhizhao Niu:
Yes, all CMOS inputs are built like this.
The two transistors indeed form an inverter. But how the internal logic interprets the signal does not matter for you; the device is disabled when the voltage at the outside, at the OE pin, is low.
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